Problem 89 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$x^{-2 / 3}+x^{-1 / 3}-6=0$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{1}{8} \).

Step 1: Substitute the variable

Let’s define a new variable to make the equation simpler. Let’s call it \( y = x^{-1/3} \). Now we need to express the original equation in terms of \( y \).

Step 2: Rewrite the equation in terms of the new variable

Substitute \( y = x^{-1/3} \) into the original equation. This gives us: \[ y^2 + y - 6 = 0 \]

Step 3: Solve the quadratic equation

Solve the quadratic equation \( y^2 + y - 6 = 0 \) by factoring. We look for two numbers that multiply to -6 and add up to 1. The numbers are 3 and -2. So, we factor the equation as: \[ (y + 3)(y - 2) = 0 \]

Step 4: Find the values of the new variable

Set each factor equal to zero and solve for \( y \): \[ y + 3 = 0 \] \[ y = -3 \] \[ y - 2 = 0 \] \[ y = 2 \]

Step 5: Substitute back the original variable

Recall that \( y = x^{-1/3} \). Now we substitute back to find the values of \( x \): For \( y = -3 \): \[ -3 = x^{-1/3} \] Raise both sides to the power of -3: \[ (-3)^{-3} = x \] \[ -\frac{1}{27} = x \] This is not valid as the result for a real number, hence it is disregarded. Next, For \( y = 2 \): \[ 2 = x^{-1/3} \] Raise both sides to the power of -3: \[2^{-3} = x \] \[ \frac{1}{8} = x \]

Key Concepts

substitution method

The substitution method is a powerful tool to simplify complex equations. Imagine juggling a complex problem; sometimes, it’s easier to rename difficult parts as something simpler. Here, we'll use substitution to rewrite parts of an equation for easier handling.

First, look at the original equation: \underline{\phantom{xxx}} \[x^{-2 / 3}+x^{-1 / 3}-6=0\]. This might look intimidating due to the fractional exponents. Hence, we substitute a part of it to simplify. We set \( y = x^{-1/3} \). Now, the original equation transforms to \( y^2 + y - 6 = 0 \).

This step of substitution is crucial as it makes a seemingly complex equation more manageable. By substituting, we reduce the equation to a familiar and simpler form, leading us to a quadratic equation.

quadratic equations

Quadratic equations are polynomials of degree 2 and appear in the form \( ax^2 + bx + c = 0 \). Here, our equation is \( y^2 + y - 6 = 0 \).

These equations often have two solutions and are at the heart of many algebra problems. There are several methods to solve quadratic equations, including factoring, completing the square, or using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In this case, factoring is used as it's the simplest route. By finding two numbers that multiply to -6 (the constant term) and add to 1 (the coefficient of the linear term), we get 3 and -2. Hence, the quadratic equation is factored as \( (y + 3)(y - 2) = 0 \).

factoring

Factoring is an essential algebraic method where an expression is rewritten as the product of its factors. This method is particularly useful for solving quadratic equations.

Taking our factored equation \( (y + 3)(y - 2) = 0 \), we set each factor equal to zero:
\begin{align*} y+3&=0 \ y&=-3 \ y-2&=0 \ y&=2 \ \text{Consequently, we find two solutions for } & y: -3 \text{ and } 2. \ \text{Remember, we initially set } y=x^{-1/3}, \text{ thus substituting our answers back gives us:} \ \text{For } & y=-3: y = x^{-1/3}, \ -3 = x^{-1/3}, \text{ when we attempt to solve this by raising both sides to the power of -3, } \text{we get non-real solutions}. \text{ Hence, } -3 \text{ is disregarded}. \ \text{For } & y=2: y = x^{-1/3}, \ 2 = x^{-1/3}, \ 2^{-3} = x, \ 1/8 = x. \ \text{The valid solution is } & x=1/8. \ Hence, the power of factoring comes in simplifying the equation and solving it step by step.}